0
$\begingroup$

In O'searcoid, Metric Spaces, he provides the following example of a metric space:

Suppose C is a circle and, for each $a,b ∈ C$, define $d(a,b)$ to be the distance along the line segment from $a$ to $b$. Then $d$ is a metric on $C$.

I have decided to confirm that his example is, indeed, a metric (here might be a good place to admit that I am weak in trigonometry).


According to wikipedia, the length of a line segment (or chord) can be written as:

$$d(a,b) = 2 r \sin{ \dfrac{ \theta }{2} }$$

where $r$ denotes the radius of $C$ and $\theta$ the angle formed between points $a,b$ with centre.

From here, it can clearly be seen that if $a = b$ then $d = 0$ since $\theta = 0$. Further, since $\theta \ge 0$, it follows that $d \ge 0$. Finally, $d(a,b) = d(b,a)$ since the angle is determined by the relation of these points with the centre of the circle. Thus, all that remains is the triangle inequality.

Take three points in $C$, $a,b,c$. The triangle inequality states that:

$$ d ( a,b ) \le d( a,c ) + d( c,b ) $$

Thus the following must satisfy:

$$ 2 r \sin{ \dfrac{ \theta_{ab} }{2} } \le 2 r \sin{ \dfrac{ \theta_{ac} }{2} } + 2 r \sin{ \dfrac{ \theta_{cb} }{2} }$$

From here, it may be deduced that $\theta_{ab} = \theta_{ac} + \theta_{cb} $ since they essentially just "split" the angle into two parts.

Removing like terms from the triangle inequality yields:

$$ \sin{ \dfrac{ \theta_{ab} }{2} } \le \sin{ \dfrac{ \theta_{ac} }{2} } + \sin{ \dfrac{ \theta_{cb} }{2} }$$

A wolfram calculation demonstrates that an alternative form for an equation in the form $ \sin{ \frac{x}{2} } + \sin{ \frac{y}{2} }$ is $2 \sin{(\frac{x}{4}+\frac{y}{4})} \cos{(\frac{x}{4}-\frac{y}{4})}$. A further wolfram calculation shows that an equation in the form $ \sin{ \frac{x}{2} }$ may be written as $2 \sin{ \frac{x}{4} } \cos{ \frac{x}{4} }$ I can rewrite the triangle inequality as:

$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{( \frac{ \theta_{bc} }{4} + \frac{ \theta_{cb}}{4} )} \cos{( \frac{ \theta_{bc} }{4} - \frac{ \theta_{cb}}{4} )} $$

And using the fact that $\theta_{ab} = \theta_{ac} + \theta_{cb} $, this can be written:

$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{( \frac{ \theta_{ab} - \theta_{cb} }{4} + \frac{ \theta_{ab} - \theta_{bc}}{4} )} \cos{( \frac{ \theta_{ab} - \theta_{cb} }{4} - \frac{ \theta_{ac} - \theta_{bc}}{4} )} $$

And this simplifies to:

$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ - \theta_{ab} } {4} } $$

And this $ \cos{-x} = \cos{x} $, we end up with:

$$ \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } \le \sin{ \dfrac{ \theta_{ab} }{4} } \cos{ \dfrac{ \theta_{ab} } {4} } $$


From what I can see, this also demonstrates that:

$$ \sin{ \frac{x}{2} } = \sin{ \frac{y}{2} } + \sin{ \frac{z}{2} } $$

where $x = y + z$ (just the original triangle inequality).

So based upon this I decided to use some test values. However, I am finding that the triangle inequality (equality) is not holding for the calculations I put in. Thus, there must be an error in the proof somewhere.

Thus my question is simply: What am I doing wrong?

$\endgroup$
  • $\begingroup$ Well, fooling around with sines is a bad idea. This is the ordinary triangle inequality. With a real triangle. $\endgroup$ – André Nicolas Jun 26 '13 at 5:12
  • $\begingroup$ @CalvinLin I thought I was using straight-line distance and not arclenth? $\endgroup$ – GovEcon Jun 26 '13 at 5:16
  • $\begingroup$ You're right. My bad. Sorry. $\endgroup$ – Calvin Lin Jun 26 '13 at 5:18
  • $\begingroup$ @AndréNicolas Thank-you for the point. I am working with triangles a circle now. For my future knowledge, could you expand on bit on the point that "fooling around with sines is a bad idea?" $\endgroup$ – GovEcon Jun 26 '13 at 5:21
  • 1
    $\begingroup$ Well, the circle is irrelevant, you are talking about ordinary Euclidean distance. Inequalities that involve trig functions of several angles can be quite messy. $\endgroup$ – André Nicolas Jun 26 '13 at 5:25
1
$\begingroup$

The first thing that you’re doing wrong is making the whole business way too complicated! These distances are identical to the ordinary straight-line Euclidean distances between the points. If you accept that the usual metric in the plane is a metric, then this $d$ must be a metric as well: it’s the same metric, just restricted to the circle $C$.

Alternatively, you can think of it in an even more elementary way. The points $a,b$, and $c$ are the vertices of a triangle $\triangle abc$. (The triangle can be degenerate if two or all three points are the same, but you can deal with those cases separately if you wish.) The distances given by the function $d$ are just the ordinary lengths of the sides of that triangle, so of course they satisfy the triangle inequality.

That said, let’s take a look at how you might verify your (correct) inequality

$$2r\sin\frac{\theta_{ab}}2\le 2r\sin\frac{\theta_{ac}}2+2r\sin\frac{\theta_{cb}}2\;.$$

First a small terminological correction: when you replace it by

$$\sin\frac{\theta_{ab}}2\le\sin\frac{\theta_{ac}}2+\sin\frac{\theta_{cb}}2\;,$$

you’re not ‘[r]emoving like terms’: you’re cancelling (or dividing out) a common factor (of $2r$). The easiest way to proceed is to write $\theta_{ab}=\theta_{ac}+\theta_{cb}$ and use the sum formula for the sine:

$$\sin\frac{\theta_{ab}}2=\sin\left(\frac{\theta_{ac}}2+\frac{\theta_{cb}}2\right)=\sin\frac{\theta_{ac}}2\cos\frac{\theta_{cb}}2+\sin\frac{\theta_{cb}}2\cos\frac{\theta_{ac}}2\;.$$

Now all cosines are between $-1$ and $1$, so

$$\cos\frac{\theta_{cb}}2\le 1\quad\text{and}\quad\cos\frac{\theta_{ac}}2\le 1\;,$$

and it immediately follows that

$$\sin\frac{\theta_{ac}}2\cos\frac{\theta_{cb}}2+\sin\frac{\theta_{cb}}2\cos\frac{\theta_{ac}}2\le\sin\frac{\theta_{ac}}2\cdot 1+\sin\frac{\theta_{cb}}2\cdot 1=\sin\frac{\theta_{ac}}2+\sin\frac{\theta_{cb}}2\;,$$

which is exactly what you want.

$\endgroup$
  • $\begingroup$ Thank you for the incredible answer - providing a simpler approach, demonstrating where I went wrong, and also making some corrections to my proof-writing. Much appreciated. :) $\endgroup$ – GovEcon Jun 26 '13 at 5:32
  • $\begingroup$ @Jordan: You’re very welcome. $\endgroup$ – Brian M. Scott Jun 26 '13 at 5:35
2
$\begingroup$

The easy way to do this is to apply to the triangle inequality directly.


Otherwise, you can use the sum formula, to get that

$\sin \frac{x}{2} = \sin \frac{y+z}{2} = \sin \frac{y}{2} \cos \frac{z}{2} + \cos \frac{y}{2} \sin \frac{z}{2} \leq \sin \frac{y}{2} + \sin \frac{z}{2} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.